(no subject)

[sigmaleph]

Check my math?

You flip a coin until it comes up tails. If you get tails immediately, you win $1. If you get one head and then lose on the second round, you win $2. If you get (n-1) heads and lose on the nth round, you win $ 2^(n-1).

Probability of you losing on the first round is 1/2. Probability of you losing on the nth round is 2^-n. Expected value of W (how much you win in USD) is therefore 1/2 + 2*1/4 + ... + 2^(n-1)*2*(^-n) + ... which is a sum of infinitely many terms all equal to 1/2. So the expected value of how much you win is infinite. Is that correct?

(this is not a puzzle or anything, I'm fairly confident I'm right but could be making a mistake)

Comments

[contrarianarchon]

Showed this to my dad, who is a mathematician, his reply quoted below

"Re the betting strategy. It is a well known idea sometimes called the martingale strategy. It is a really good example of convergence problems in probability which are complex. In this case the flaw (as a practical betting strategy if you imagine doubling your bets each time you lose) is that you need an infinite reservoir of funds. Any cap, however large, breaks it."

Which tl:drs to "yes"

[sigmaleph]

Thank you! I was certainly not thinking of it as a betting strategy, more as an unusually generous game someone might offer you (in which case, other people have told me, it's called the St Petersburg lottery).